Question: In a class of $9$, there are $4$ students who have done their homework. If the teacher chooses $3$ students, what is the probability that none of the three of them have done their homework?
Answer: We can think about this problem as the probability of $3$ events happening. The first event is the teacher choosing one student who has not done his homework. The second event is the teacher choosing another student who has not done his homework, given that the teacher already chose someone who has not done his homework, and so on. The probabilty that the teacher will choose someone who has not done his homework is the number of students who have not done their homework divided by the total number of students: $\dfrac{5} {9}$ Once the teacher's chosen one student, there are only $8$ left. There's also one fewer student who has not done his homework, since the teacher isn't going to pick the same student twice. So, the probability that the teacher picks a second student who also has not done his homework is $\dfrac{4} {8}$ The probability of the teacher picking two students who have not done their homework must then be $\dfrac{5} {9} \cdot \dfrac{4} {8}$ We can continue using the same logic for the rest of the students the teacher picks. So, the probability of the teacher picking $3$ students such that none of them have done their homework is $\dfrac{5}{9}\cdot\dfrac{4}{8}\cdot\dfrac{3}{7} = \dfrac{5}{42}$